package leetcode.editor.cn;
//给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。 
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// 叶子节点 是指没有子节点的节点。 
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// 示例 1： 
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//输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
//输出：[[5,4,11,2],[5,8,4,5]]
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// 示例 2： 
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//输入：root = [1,2,3], targetSum = 5
//输出：[]
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// 示例 3： 
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//输入：root = [1,2], targetSum = 0
//输出：[]
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// 提示： 
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// 树中节点总数在范围 [0, 5000] 内 
// -1000 <= Node.val <= 1000 
// -1000 <= targetSum <= 1000 
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// Related Topics 树 深度优先搜索 回溯 二叉树 👍 1072 👎 0


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import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution265 {
    List<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> trace = new LinkedList<>();
    int traceSum = 0;

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        dfs(root, targetSum);
        return res;
    }

    private void dfs(TreeNode node, int targetSum) {
        if (node == null) return;

        trace.add(node.val);
        traceSum += node.val;
        if (traceSum == targetSum && node.left == null && node.right == null) {
            res.add(new LinkedList<>(trace));
            trace.removeLast();
            traceSum -= node.val;
            return;
        }
        dfs(node.left, targetSum);
        dfs(node.right, targetSum);

        trace.removeLast();
        traceSum -= node.val;
    }

    public static void main(String[] args) {
        TreeNode _4 = new TreeNode(4);
        TreeNode __4 = new TreeNode(-4);
        TreeNode _8 = new TreeNode(8, null, __4);
        TreeNode _5 = new TreeNode(5, _4, _8);
        Solution265 s = new Solution265();
        System.out.println(s.pathSum(_5, 9));
    }

    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {
        }

        TreeNode(int val) {
            this.val = val;
        }

        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }
}
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